Integrand size = 23, antiderivative size = 113 \[ \int x^8 \sqrt {\frac {1-x^3}{1+x^3}} \, dx=\frac {1}{2} \sqrt {\frac {1-x^3}{1+x^3}} \left (1+x^3\right )-\frac {1}{6} \sqrt {\frac {1-x^3}{1+x^3}} \left (1+x^3\right )^2-\frac {1}{9} \left (\frac {1-x^3}{1+x^3}\right )^{3/2} \left (1+x^3\right )^3-\frac {1}{3} \arctan \left (\sqrt {\frac {1-x^3}{1+x^3}}\right ) \]
-1/9*((-x^3+1)/(x^3+1))^(3/2)*(x^3+1)^3-1/3*arctan(((-x^3+1)/(x^3+1))^(1/2 ))+1/2*(x^3+1)*((-x^3+1)/(x^3+1))^(1/2)-1/6*(x^3+1)^2*((-x^3+1)/(x^3+1))^( 1/2)
Time = 0.47 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.84 \[ \int x^8 \sqrt {\frac {1-x^3}{1+x^3}} \, dx=\frac {\sqrt {\frac {1-x^3}{1+x^3}} \left (\sqrt {1-x^3} \left (4+x^3-x^6+2 x^9\right )-6 \sqrt {1+x^3} \arctan \left (\frac {\sqrt {1-x^3}}{\sqrt {1+x^3}}\right )\right )}{18 \sqrt {1-x^3}} \]
(Sqrt[(1 - x^3)/(1 + x^3)]*(Sqrt[1 - x^3]*(4 + x^3 - x^6 + 2*x^9) - 6*Sqrt [1 + x^3]*ArcTan[Sqrt[1 - x^3]/Sqrt[1 + x^3]]))/(18*Sqrt[1 - x^3])
Time = 0.24 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.96, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {2053, 2052, 366, 27, 360, 27, 298, 216}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^8 \sqrt {\frac {1-x^3}{x^3+1}} \, dx\) |
\(\Big \downarrow \) 2053 |
\(\displaystyle \frac {1}{3} \int x^6 \sqrt {\frac {1-x^3}{x^3+1}}dx^3\) |
\(\Big \downarrow \) 2052 |
\(\displaystyle -\frac {4}{3} \int \frac {x^6 \left (1-x^6\right )^2}{\left (x^6+1\right )^4}d\sqrt {\frac {1-x^3}{x^3+1}}\) |
\(\Big \downarrow \) 366 |
\(\displaystyle -\frac {4}{3} \left (\frac {2 x^9}{3 \left (x^6+1\right )^3}-\frac {1}{6} \int \frac {6 x^6 \left (1-x^6\right )}{\left (x^6+1\right )^3}d\sqrt {\frac {1-x^3}{x^3+1}}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {4}{3} \left (\frac {2 x^9}{3 \left (x^6+1\right )^3}-\int \frac {x^6 \left (1-x^6\right )}{\left (x^6+1\right )^3}d\sqrt {\frac {1-x^3}{x^3+1}}\right )\) |
\(\Big \downarrow \) 360 |
\(\displaystyle -\frac {4}{3} \left (\frac {1}{4} \int -\frac {2 \left (1-2 x^6\right )}{\left (x^6+1\right )^2}d\sqrt {\frac {1-x^3}{x^3+1}}+\frac {2 x^9}{3 \left (x^6+1\right )^3}+\frac {\sqrt {\frac {1-x^3}{x^3+1}}}{2 \left (x^6+1\right )^2}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {4}{3} \left (-\frac {1}{2} \int \frac {1-2 x^6}{\left (x^6+1\right )^2}d\sqrt {\frac {1-x^3}{x^3+1}}+\frac {2 x^9}{3 \left (x^6+1\right )^3}+\frac {\sqrt {\frac {1-x^3}{x^3+1}}}{2 \left (x^6+1\right )^2}\right )\) |
\(\Big \downarrow \) 298 |
\(\displaystyle -\frac {4}{3} \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{x^6+1}d\sqrt {\frac {1-x^3}{x^3+1}}-\frac {3 \sqrt {\frac {1-x^3}{x^3+1}}}{2 \left (x^6+1\right )}\right )+\frac {2 x^9}{3 \left (x^6+1\right )^3}+\frac {\sqrt {\frac {1-x^3}{x^3+1}}}{2 \left (x^6+1\right )^2}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {4}{3} \left (\frac {1}{2} \left (\frac {1}{2} \arctan \left (\sqrt {\frac {1-x^3}{x^3+1}}\right )-\frac {3 \sqrt {\frac {1-x^3}{x^3+1}}}{2 \left (x^6+1\right )}\right )+\frac {2 x^9}{3 \left (x^6+1\right )^3}+\frac {\sqrt {\frac {1-x^3}{x^3+1}}}{2 \left (x^6+1\right )^2}\right )\) |
(-4*((2*x^9)/(3*(1 + x^6)^3) + Sqrt[(1 - x^3)/(1 + x^3)]/(2*(1 + x^6)^2) + ((-3*Sqrt[(1 - x^3)/(1 + x^3)])/(2*(1 + x^6)) + ArcTan[Sqrt[(1 - x^3)/(1 + x^3)]]/2)/2))/3
3.3.92.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 2*p + 3))/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] : > Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Simp[1/(2*b^(m/2 + 1)*(p + 1)) Int[(a + b*x^2)^(p + 1)*Expan dToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)] - (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[m/2, 0] & & (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^2, x_Symbol] :> Simp[(-(b*c - a*d)^2)*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(2*a* b^2*e*(p + 1))), x] + Simp[1/(2*a*b^2*(p + 1)) Int[(e*x)^m*(a + b*x^2)^(p + 1)*Simp[(b*c - a*d)^2*(m + 1) + 2*b^2*c^2*(p + 1) + 2*a*b*d^2*(p + 1)*x^ 2, x], x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && LtQ[p , -1]
Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)))/((c_) + (d_.)*(x_)))^(p_), x_S ymbol] :> With[{q = Denominator[p]}, Simp[q*e*(b*c - a*d) Subst[Int[x^(q* (p + 1) - 1)*(((-a)*e + c*x^q)^m/(b*e - d*x^q)^(m + 2)), x], x, (e*((a + b* x)/(c + d*x)))^(1/q)], x]] /; FreeQ[{a, b, c, d, e, m}, x] && FractionQ[p] && IntegerQ[m]
Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.) ))^(p_), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(e*( (a + b*x)/(c + d*x)))^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Time = 0.87 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.71
method | result | size |
risch | \(\frac {\left (x^{3}+1\right ) \left (2 x^{6}-3 x^{3}+4\right ) \sqrt {-\frac {x^{3}-1}{x^{3}+1}}}{18}-\frac {\arcsin \left (x^{3}\right ) \sqrt {-\frac {x^{3}-1}{x^{3}+1}}\, \sqrt {-\left (x^{3}+1\right ) \left (x^{3}-1\right )}}{6 \left (x^{3}-1\right )}\) | \(80\) |
trager | \(\frac {\left (x^{3}+1\right ) \left (2 x^{6}-3 x^{3}+4\right ) \sqrt {-\frac {x^{3}-1}{x^{3}+1}}}{18}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {-\frac {x^{3}-1}{x^{3}+1}}\, x^{3}+x^{3}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {-\frac {x^{3}-1}{x^{3}+1}}\right )}{6}\) | \(99\) |
1/18*(x^3+1)*(2*x^6-3*x^3+4)*(-(x^3-1)/(x^3+1))^(1/2)-1/6*arcsin(x^3)*(-(x ^3-1)/(x^3+1))^(1/2)*(-(x^3+1)*(x^3-1))^(1/2)/(x^3-1)
Time = 0.34 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.58 \[ \int x^8 \sqrt {\frac {1-x^3}{1+x^3}} \, dx=\frac {1}{18} \, {\left (2 \, x^{9} - x^{6} + x^{3} + 4\right )} \sqrt {-\frac {x^{3} - 1}{x^{3} + 1}} - \frac {1}{3} \, \arctan \left (\frac {{\left (x^{3} + 1\right )} \sqrt {-\frac {x^{3} - 1}{x^{3} + 1}} - 1}{x^{3}}\right ) \]
1/18*(2*x^9 - x^6 + x^3 + 4)*sqrt(-(x^3 - 1)/(x^3 + 1)) - 1/3*arctan(((x^3 + 1)*sqrt(-(x^3 - 1)/(x^3 + 1)) - 1)/x^3)
Timed out. \[ \int x^8 \sqrt {\frac {1-x^3}{1+x^3}} \, dx=\text {Timed out} \]
\[ \int x^8 \sqrt {\frac {1-x^3}{1+x^3}} \, dx=\int { x^{8} \sqrt {-\frac {x^{3} - 1}{x^{3} + 1}} \,d x } \]
Time = 0.32 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.50 \[ \int x^8 \sqrt {\frac {1-x^3}{1+x^3}} \, dx=\frac {1}{6} \, \arcsin \left (x^{3}\right ) \mathrm {sgn}\left (x^{3} + 1\right ) + \frac {1}{18} \, \sqrt {-x^{6} + 1} {\left ({\left (2 \, x^{3} \mathrm {sgn}\left (x^{3} + 1\right ) - 3 \, \mathrm {sgn}\left (x^{3} + 1\right )\right )} x^{3} + 4 \, \mathrm {sgn}\left (x^{3} + 1\right )\right )} \]
1/6*arcsin(x^3)*sgn(x^3 + 1) + 1/18*sqrt(-x^6 + 1)*((2*x^3*sgn(x^3 + 1) - 3*sgn(x^3 + 1))*x^3 + 4*sgn(x^3 + 1))
Time = 16.90 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.89 \[ \int x^8 \sqrt {\frac {1-x^3}{1+x^3}} \, dx=\frac {2\,\sqrt {-\frac {x^3-1}{x^3+1}}}{9}-\frac {\mathrm {atan}\left (\sqrt {-\frac {x^3-1}{x^3+1}}\right )}{3}+\frac {x^3\,\sqrt {-\frac {x^3-1}{x^3+1}}}{18}-\frac {x^6\,\sqrt {-\frac {x^3-1}{x^3+1}}}{18}+\frac {x^9\,\sqrt {-\frac {x^3-1}{x^3+1}}}{9} \]